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**Topic**

Given two straight lines_{first}: *mx* – 2*y* – 1 = 0 and_{2}: *x* – 2*y* + 3 = 0. With what value of parameter *m* then:

a)_{first} //_{2}?

b)_{first}\( \bot {\Delta _2}\)?

Given 2 straight lines_{first}: *ax* + *by* + *c* = 0 and_{2}: *a’x* + *b’y* + *c’* = 0. We have_{first} //_{2 }\( \Leftrightarrow \frac{a}{{a’}} = \frac{b}{{b’}} \ne \frac{c}{{c’}}\)

Step 1: Apply the above results to find *m* satisfied_{first} //_{2}

Step 2: Find *m* to 2 VTPT of_{first} and_{2 }multiplying scalars together by 0 satisfy ∆_{first}\( \bot {\Delta _2}\)

**Detailed explanation**

∆_{first }whose VTPT is \(\overrightarrow {{n_1}} = (m; – 2)\); ∆_{2 }whose VTPT is \(\overrightarrow {{n_2}} = (1; – 2)\)

a)_{first} //_{2 }if and only if \(\overrightarrow {{n_1}} \) and \(\overrightarrow {{n_2}} \) are in the same direction and ∆_{first }and_{2} not match

\( \Leftrightarrow \frac{m}{1} = \frac{{ – 2}}{{ – 2}} \ne \frac{{ – 1}}{3} \Leftrightarrow m = 1\)

So with *m* = 1 then_{first} //_{2}

b) \({\Delta _1} \bot {\Delta _2} \Leftrightarrow \overrightarrow {{n_1}} .\overrightarrow {{n_2}} = 0 \Leftrightarrow m + 4 = 0 \Leftrightarrow m = – 4\)

So with *m* = -4 then \({\Delta _1} \bot {\Delta _2}\)

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